In the 1989, Journal of Number Theory paper, N. Tzanakis and B. de Wegner showed that the largest of the 6 triangular numbers that are products of three consecutive integers is 258,474,216. What are the three consecutive integers? Solution: Let x represent the first consecutive integer, x+1 the second consecutive integer and x+2 the third consecutive integer, then x(x+1)(x+2) = 258,474,216 x(x² + 3x+2) = 258,474,216 x+3x²+2x=258,474,216 x+3x+2x-258, 474,216 = 0 Use the factor theorem, long division and the quadratic formula 4
In the 1989, Journal of Number Theory paper, N. Tzanakis and B. de Wegner showed that the largest of the 6 triangular numbers that are products of three consecutive integers is 258,474,216. What are the three consecutive integers? Solution: Let x represent the first consecutive integer, x+1 the second consecutive integer and x+2 the third consecutive integer, then x(x+1)(x+2) = 258,474,216 x(x² + 3x+2) = 258,474,216 x+3x²+2x=258,474,216 x+3x+2x-258, 474,216 = 0 Use the factor theorem, long division and the quadratic formula 4
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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### Finding Consecutive Integers that are the Product of Three Consecutive Integers
In the 1989 Journal of Number Theory paper, N. Tzankis and B. deWegner showed that the largest of the six triangular numbers that are products of three consecutive integers is 258,474,216. The challenge is to determine these three consecutive integers.
#### Problem:
What are the three consecutive integers?
#### Solution:
Let \( x \) represent the first consecutive integer, \( x+1 \) the second consecutive integer, and \( x+2 \) the third consecutive integer.
Then we write the equation:
\[ x(x+1)(x+2) = 258,474,216 \]
Expanding the left-hand side, we get:
\[ x(x^2 + 3x + 2) = 258,474,216 \]
\[ x^3 + 3x^2 + 2x = 258,474,216 \]
\[ x^3 + 3x^2 + 2x - 258,474,216 = 0 \]
This is a cubic equation which can be solved using the factor theorem, long division, and the quadratic formula.
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Transcribed Image Text:---
### Finding Consecutive Integers that are the Product of Three Consecutive Integers
In the 1989 Journal of Number Theory paper, N. Tzankis and B. deWegner showed that the largest of the six triangular numbers that are products of three consecutive integers is 258,474,216. The challenge is to determine these three consecutive integers.
#### Problem:
What are the three consecutive integers?
#### Solution:
Let \( x \) represent the first consecutive integer, \( x+1 \) the second consecutive integer, and \( x+2 \) the third consecutive integer.
Then we write the equation:
\[ x(x+1)(x+2) = 258,474,216 \]
Expanding the left-hand side, we get:
\[ x(x^2 + 3x + 2) = 258,474,216 \]
\[ x^3 + 3x^2 + 2x = 258,474,216 \]
\[ x^3 + 3x^2 + 2x - 258,474,216 = 0 \]
This is a cubic equation which can be solved using the factor theorem, long division, and the quadratic formula.
---
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